$0.03+$0.04=$0.15x4=$0.60
654,035 654,035 654,035 654,035 654,035 654,035 654,035
Answer:
Just getting my points back don’t mind me
Step-by-step explanation:
So... the radiator has 15 liters of 70% antifreeze.. but needs an 80% antifreeze
well, so, you need to drain some and put some with higher percentage, seems to be, you will end up at the same 15 liters, possible the radiator's capacity, of 80% antifreeze
so, the same amount going out, of 70% is the same amount going in, of 100% antifreeze
now.. let's use the decimal format for the percents, or 70% is 70/100 or 0.7 and so on

so.. let's subtract, from the current solution, 0.7x and add 1x or x, our antifreeze concentration amount, should be 12 though
10.5 - 0.7x + x = 12
solve for "x"
Answer:
(1, 3)
Step-by-step explanation:
Current position (3,2)
Moving left of the x axis results in a decrease, so: 3 - 2 (As she is moving 2 units to the left)
Moving up results in a positive increase on the y axis, so: 2 + 1 (As she is moving one unit up)